1.

o is a point in the interior of triangle ABC, OD is perpendicular on BC, OE is perpendicular on AC, AND OF is perpendicular on AB prove that AF×BD×CE=AE×BF×CD​

Answer»

ong>Answer:

Let O be a POINT in the interior of △ABC, and let OD⊥BC,OE⊥CA and OF⊥AB

(i) In right triangles △OFA,△ODB and △OEC we have

OA

2

=AF

2

+OF

2

OB

2

=BD

2

+OD

2

and, OC

2

=CE

2

+OE

2

Adding all these results, we GET

OA

2

+OB

2

+OC

2

=AF

2

+BD

2

+CE

2

+OF

2

+OD

2

+OE

2

⇒ AF

2

+BD

2

+CE

2

=OA

2

+OB

2

+OC

2

−OD

2

−OE

2

−OF

2

[HENCE PROVED]

(II) In right triangles △ODB and △ODC, we have

OB

2

=OD

2

+BD

2


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